ch 7 3 using chemical formulas
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Ch 7.3 Using Chemical Formulas. The Mass of a Mole of an Element. Remember: The atomic mass of an element (a single atom) is expressed in atomic mass units (amu). Molar Mass: is the atomic mass of an element expressed in grams/mole (g/mol). Carbon = 12.01 g/mol Hydrogen = 1.01 g/mol

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the mass of a mole of an element
The Mass of a Mole of an Element
  • Remember: The atomic mass of an element (a single atom) is expressed in atomic mass units (amu).
  • Molar Mass: is the atomic mass of an element expressed in grams/mole (g/mol).
  • Carbon = 12.01 g/mol
  • Hydrogen = 1.01 g/mol
  • When dealing with molar mass, round off to two decimals. 12.011g/mol -> 12.01g/mol
the mass of a mole of a compound
The Mass of a Mole of a Compound
  • You calculate the mass of a molecule by adding up the molar masses of the atoms making up the molecules.
  • Example: H2O
    • H = 1.01 g x 2 atoms = 2.02 g/mol
    • O = 16.00 g x 1 atom = 16.00 g/mol
  • Molar Mass of H2O = 2.02 g/mol + 16.00 g/mol = 18.02 g/mol
  • This applies to both molecular and ionic compounds
slide4
Find the molar mass of PCl3
    • P = 30.97 g x 1 atom = 30.97 g/mol
    • Cl = 35.45 g x 3 atoms = 106.35 g/mol
    • PCl3 = 30.97 g + 106.35 g = 137.32 g/mol
  • What is the molar mass of Sodium Hydrogen Carbonate (NaHCO3) ?
    • Na = 22.99 g x 1 atom = 22.99 g/mol
    • H = 1.01 g x 1 atom = 1.01 g/mol
    • C = 12.01 g x 1 atom = 12.01 g/mol
    • O = 16.00 g x 3 atoms = 48.00 g/mol
    • NaHCO3 = 22.99 + 1.01 + 12.01 + 48.00

= 84.01 g/mol

converting moles to mass
Converting Moles to Mass
  • You can use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance.
  • Mass (g) = # of moles x mass (g)

1 mole

Example: If molar mass of NaCl is 58.44 g/mol, what is the mass of 3.00 mol NaCl?

Mass of NaCl = 3.00 mol x 58.44g =

1 mol

175 g NaCl

example 2 moles to mass
Example 2: Moles to Mass
  • What is the mass of 9.45 mol of Aluminum Oxide (Al2O3)?
  • Find molar mass of Al2O3

= 101.96 g/mol

  • Mass = 9.45 mol Al2O3 x 101.96 g Al2O3

1 mol Al2O3

= 964 g Al2O3

converting mass to moles
Converting Mass to Moles
  • You can invert the conversion factor to find moles when given the mass.
  • Moles = mass (g) x 1 mole

mass (g)

Example: If molar mass of Na2SO4 142.05 g/mol, how many moles is 10.0 g of Na2SO4?

Moles of Na2SO4 = 10.0 g x 1 mol =

142.05 g

= 0.0704 mol Na2SO4

example 2 mass to moles
Example 2: Mass to Moles
  • How many moles are in 75.0 g of Dinitrogen Trioxide?
  • Find molar mass of N2O3

= 76.02 g/mol

  • Moles = 75.0 g N2O3 x 1 mole =

76.02 g

N2O3

0.987 mol N2O3

percent composition
Percent Composition
  • Percent Composition: the relative amount of the elements in a compound.
  • Also known as the percent by mass
  • It can be calculated in two ways:
    • Using Mass Data
    • Using the Chemical Formula

% mass of element= mass of element x100% mass of compound

example
Example
  • When a 13.60 g sample of a compound containing Mg and O is decomposed, 5.40 g O is obtained. What is the % composition of this compound?

Mass of compound: 13.60 g

Mass of oxygen: 5.40 g O

Mass of magnesium: 13.60 g - 5.40 g = 8.20 g Mg

% Mg = 8.20 g Mg x 100% =

13.60 g

% O = 5.40 g O x 100% =

13.60 g

60.3%

39.7%

slide11
Find the percent composition of Cu2S.
  • Find mass of Cu and S
    • Cu = 63.55 x 2 = 127.10 g
    • S = 32.07 g
  • Find mass of Cu2S
    • 127.10 g + 32.07 g = 159.17 g

% Composition

    • Cu = 127.10 g x 100% =

159.17 g

    • S = 32.07 g x 100% =

159.17 g

79.85%

20.15%

homework
Homework
  • 7.3 pg 253 #30-33
empirical formulas
Empirical Formulas
  • Empirical Formula: shows the smallest whole-number ratio of the atoms of the elements in a compound.
  • Example:
    • The Empirical Formula for Hydrogen Peroxide (H2O2) is HO with a 1:1 ratio.
    • The Empirical Formula for Carbon Dioxide (CO2) is CO2 with a 1:2 ratio.
determining the empirical formula of a compound
Determining the Empirical Formula of a Compound
  • A compound is found to contain 25.9% Nitrogen and 74.1% Oxygen. What is the Empirical Formula of the compound?
  • 25.9 g N x 1 mol N = 14.01 g N
  • 74.1 g O x 1 mol O = 16.00 g O
  • N1.85O4.63 = N1O2.5 = N2O5

1.85 mol N

4.63 mol O

molecular formulas
Molecular Formulas
  • Molecular Formula: tells the actual number of each kind of atom present in a molecule of a compound
  • Example:
    • The Molecular Formula for Hydrogen Peroxide is H2O2.
    • The Molecular Formula for Carbon Dioxide is CO2
  • It is possible to find the Molecular Formula using the Empirical Formula if you know the molar mass of the compound.
finding the molecular formula
Finding the Molecular Formula
  • Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N
  • Step 1: Find the empirical formula molar mass
    • 12.01 + (4 x 1.01) + 14.01 = 30.06 g/mol
  • Step 2: Divide molar mass by EF molar mass
    • 60.0 g/mol = 1.99  2 30.06 g/mol
  • Step 3: Multiply empirical formula by 2
    • CH4N x 2 = C2H8N2
homework1
Homework
  • 7.4 pg 253 #36-38
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